1gm of HCI and 1gm NaOH are dissolved together in one litre solution of water. The resulting solution is:?​

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Explanation:

spammer op

As we know that,

no. of moles=

Mol. wt.

Wt.

Mol. wt. of NaOH=40g

Wt. of NaOH=2g(Given)

Therefore,

No. of mles =

40

2

=0.05 mol

As we know that molarity of a solution is defined as the no. of moles of solute present in 1L of solution.

Since 0.05 mole of NaOH are pesent in 1L of of solution.

Therefore,

Molarity of solution (M)=0.05

⇒[OH

−

]=0.05

As we know that,

pOH=−log[OH

−

]

⇒pOH=−log(5×10

−2

)

⇒pOH=1.3

Therefore,

pH+pOH=1

⇒pH=14−pOH=14−1.3=12.7

Hence pH of the solution is 12.7.

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when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol